2 Improved text and content edited Sep 3 '15 at 20:34 Prish Chakraborty 9011 silver badge99 bronze badges According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, however the situation asks us to consider an ideal pulley (no rotational friction at all), which implies that a majority of the momentum transferred across the pulley's string affects the mass on the other end, and the contributions towards the ceiling beingare negligible so we shouldn't care (EDIT: Refer to comments). So when we write down the equations, $$\Delta p_{i}=\Delta p_{f}$$ this should imply: $$mv\approx mV+mV+mV$$ which gives us the required result. Of course, since the pulley is ideal, (theoretically) the error of measurement would be really little. According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, however the situation asks us to consider an ideal pulley (no rotational friction at all), which implies that a majority of the momentum transferred across the pulley's string affects the mass on the other end, the contributions towards the ceiling being negligible so we shouldn't care. So when we write down the equations, $$\Delta p_{i}=\Delta p_{f}$$ this should imply: $$mv\approx mV+mV+mV$$ which gives us the required result. Of course, since the pulley is ideal, (theoretically) the error of measurement would be really little. According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, however the situation asks us to consider an ideal pulley (no rotational friction at all), which implies that a majority of the momentum transferred across the pulley's string affects the mass on the other end and the contributions towards the ceiling are negligible so we shouldn't care (EDIT: Refer to comments). So when we write down the equations, $$\Delta p_{i}=\Delta p_{f}$$ this should imply: $$mv\approx mV+mV+mV$$ which gives us the required result. Of course, since the pulley is ideal, (theoretically) the error of measurement would be really little. 1 answered Sep 3 '15 at 15:38 Prish Chakraborty 9011 silver badge99 bronze badges According to me, the momentum of the mass on the other side shouldn't be negative because the system considered is a connected system, i.e, the momentum transfer to the mass hanging on the other side of the pulley is momentum transferred through the pulley's string. I agree with @BowlOFRed about the contribution of the momentum transferred to the ceiling, however the situation asks us to consider an ideal pulley (no rotational friction at all), which implies that a majority of the momentum transferred across the pulley's string affects the mass on the other end, the contributions towards the ceiling being negligible so we shouldn't care. So when we write down the equations, $$\Delta p_{i}=\Delta p_{f}$$ this should imply: $$mv\approx mV+mV+mV$$ which gives us the required result. Of course, since the pulley is ideal, (theoretically) the error of measurement would be really little.